Problem: In the trapezoid shown, the ratio of the area of triangle $ABC$ to the area of triangle $ADC$ is $7:3$. If $AB + CD = 210$ cm, how long is segment $\overline{AB}$? [asy]
import olympiad; size(150); defaultpen(linewidth(0.8));
pair A = (0,0), B = (5,0), C = (3,2), D = (1,2);
draw(A--B--C--D--cycle--C);
label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW);
[/asy]
Solution: Let $h$ be the height of the trapezoid.  The height of the trapezoid is also a height of $ABC$ and of $ADC$.  Specifically, we have $[ABC] = (AB)(h)/2$ and $[ADC] = (CD)(h)/2$, so $[ABC]:[ADC] = AB:CD$.  Since we are given that this area ratio equals $7:3$, we know that $AB:CD = 7:3$.  Therefore, $AB = 7x$ and $CD = 3x$ for some  value of $x$.  Since $AB + CD = 210$ cm, we have $7x+3x=210$, so $10x=210$ and $x=21$. Therefore,  $AB=7 \times 21 = \boxed{147\text{ cm}}$.